We have the following maps on a complex vector space $V$
$\phi : V \rightarrow \mathbb{C}$ and $g : V^2 \rightarrow \mathbb{C}$
where $\lambda \in \mathbb{C} , x,y,w \in V$.
$\phi $ satisfies that $\phi(\lambda x ) = |\lambda|^2 \phi(x)$ and $\phi(x+y) + \phi(x-y) = 2(\phi(x) + \phi(y))$
and the map $g$ that is defined by $\phi$ via the polarization identity:
$$g(x,y) = \frac{1}{4} ( \phi(x+y) - \phi(x-y) + i \phi(x+iy) - i \phi(x-iy)$$
Now I want to show that $g$ is linear, but so far I was only able to show that $g(-x,y) = -g(x,y)$ and $g(x+y,w)= g(x,w) + g(y,w)$. But what I want to show is that $g(\lambda x,y) = \lambda g(x,y)$. (The other coordinate would be pretty much the same, so I just consider the first one.) Does anybody have an idea how this could work?